Question

Need help on my homework. I've done a problem like this before but with a simpler expression in the parenthesis. I am not sure how to even approach the problem. The question is:
Draw a right triangle to simplify the given expression. Assume x > 0.

sin(2cos^1x). (Hint: use sin2(theta) = 2sin(theta)cos(theta).

Answer 1

Was that meant to be cos^(-1)x inside?
So the inverse function of cosine? arccos(x)

Answer 2

Yes. Sorry for the confusion

Answer 3

Remember that the trig functions do something kind of interesting.
They take an angle...
and then the trig "function" applied to this angle gives a relationship based on sides of a triangle.
\[\large\rm \cos \theta=\frac{adjacent}{hyptenuse}\]

Answer 4

Inverse cosine does the opposite,
it takes a value or ratio of sides,
and spits out an angle.\[\large\rm \color{orangered}{\cos^{-1}x=\theta}\]

Answer 5

It is actually 2cos^(-1)x inside the parentheses. Is there some way to factor out the 2?

Answer 6

We'll worry about that later :)

Answer 7

We would like to do some triangle business first.

Answer 8

So we're saying that our inverse cosine function is equivalent to some angle theta,\[\large\rm \color{orangered}{\cos^{-1}x=\theta}\]Do you understand how to get from there,
\[\large\rm \cos \theta=x\]to here?

Answer 9

Yes I just worked it out and rearranged it

Answer 10

So anyway, this cos(theta)=x we can write as,\[\large\rm \cos \theta=\frac{x}{1}=\frac{adjacent}{hypotenuse}\]And we can draw a right triangle to illustrate this relationship.